TABLE OF CONTENTS We have learned that the power absorbed by a resistor can be found from p R = i 2R and this power is transformed into heat. In a long length of a conductive material, such as copper, this lost power is known as i 2R loss and thus the energy received by the load is equal to the energy transmitted minus the i 2R loss. Accordingly, we define efficiency ? as
The efficiency ? is normally expressed as a percentage. Thus,
Obviously, a good efficiency should be close to 100%
In a two-story industrial building, the total load on the first floor draws an average of 60 amperes during peak activity, while the total load of the second floor draws 40 amperes at the same time. The building receives its electric power from a 480 V source. Assuming that the total resistance of the cables (copper conductors) on the first floor is 1 ? and on the second floor is 1.6 ?, compute the efficiency of transmission.
Solution:
First, we draw a circuit that represents the electrical system of this building. This is shown in Figure 3.68.
Power p S supplied by the source:
Power loss between source and 1st floor load:
Power loss between source and 2nd floor load:
Total power loss:
Total power p L received by 1st and 2nd floor loads:
